A058067, multiplicative in a complicated way

David Wilson davidwwilson at comcast.net
Tue Jun 14 00:00:55 CEST 2005


Thank you for your message on A059067 and A046699.  With that hint, I think 
I have cracked A059067, in the sense that I now have an computationally 
efficient multiplicative formula.

Let p be prime.  Construct sequence f_p by writing each positive number k+1 
times, where k is exponent of the largest power of p dividing n.  Let g_p be 
the running sum of this sequence, starting with g_p(0) = 0.

For example,

f_2 = 1,2,2,3,4,4,4,5,6,6,7,8,8,8,8,9,10,10,11,12,12,12,...
g_2 = 0,1,3,5,8,12,16,20,25,31,37,...

f_2 is essentially A046699.  Similarly

f_3 = 1,2,3,3,4,5,6,6,7,8,9,9,9,10,11,12,12,13,14,15,15,...
g_3 = 0,1,3,6,9,13,18,24,30,37,45,...

Then it appears that A059067 is multiplicative with a(p^e) = p^(p*g_p(e)).

I have checked that with respect to the first 100 elements of g_2, g_3 and 
g_5, the multiplicative and explicit formulae for A059067 agree.  Given that 
and certain expected behaviors of the A059067 formula, I am fairly confident 
that I have the multiplicative formula right. 






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